A) Stationary with respect to the earth
B) Approaching the earth with velocity of light
C) Receding from the earth with the velocity of light
D) Receding from the earth with a velocity equal to \[1.5\times {{10}^{6}}m/s\]
Correct Answer: D
Solution :
\[\frac{\Delta \lambda }{\lambda }=\frac{v}{c},\]Now \[\Delta \lambda =\frac{0.5}{100}\lambda \Rightarrow \frac{\Delta \lambda }{\lambda }=\frac{0.5}{100}\] \[\therefore v=\frac{0.5}{100}\times c=\frac{0.5}{100}\times 3\times {{10}^{8}}=1.5\times {{10}^{6}}m/s\] Increase in l indicates that the star is receding.
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